Mathematical induction-inequations

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Johulus

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I need some help here. I don't understand at all how to solve these kind of tasks. The inequality sign completely perplexes me and I don't know what to do.

For example, \(\displaystyle 4^{n} > n^{2} \) . What should I do here? If I continue as usual; base of induction.....

\(\displaystyle n=1 \\ 4>1 \\ let's \, assume\, for \, n=k \\ 4^{k}>k^{2} \\ now \, if \, n=k+1 \\ 4^{k+1}>(k+1)^{2} \)

And what should I now do with it. I can't replace anything like in equations or anything since these are inequations. I don't understand what could I do here.
 
Well, the first thing I note is that the inequality you posted is simply not true. It doesn't work for all values of n (try n=-1). You'll need to determine bounds for which it is true in order to complete your proof.

In any case, quite often to prove one equation (or inequality) you'll end up needing to prove another one first. What if we rewrite the inequality:

\(\displaystyle 4^{k+1}>\left(k+1\right)^2\)

\(\displaystyle 4\cdot 4^k>\left(k+1\right)^2\)

\(\displaystyle 4\cdot 4^k>k^2+2k+1\)

Now, if the above is true, then that means this is also true:

\(\displaystyle 4\cdot 4^k-\left(k^2+2k+1\right)>0\)

Now, notice you've assumed that this inequality holds true for n=k:

\(\displaystyle 4^k>k^2\)

So logically this statement will also be true:

\(\displaystyle 4\cdot 4^k-\left(k^2+2k+1\right)>4k^2-\left(k^2+2k+1\right)\)

\(\displaystyle 4\cdot 4^k-\left(k^2+2k+1\right)>4k^2-k^2-2k-1\)

\(\displaystyle 4\cdot 4^k-\left(k^2+2k+1\right)>3k^2-2k-1\)

Now it looks like you're stuck. But what if we assume that \(\displaystyle 3k^2>(2k-1)\)? If that were true, then the above would resolve to:

\(\displaystyle 4\cdot 4^k-\left(k^2+2k+1\right)>0\)

Which in turn is the proof we were looking for. So you just need to prove the assumption from earlier. If you can prove \(\displaystyle 3k^2>(2k-1)\) for some set of values of n, then you've proven \(\displaystyle 4^n>n^2\) for that same set of values.
 
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